if it takes 31,500 joules of heat to warm 750g of water, what was the temperature change?

Question

My answer:

Q = (M) (DeltaT) (c)
Delta T=Q
Delta T = 31,500J
--------
750gx 1.0cal/gCo

Delta T = 42oC

Answer 1

1 calorie/gram °C = 4.186 joule/gram °C

you should use 4.186J/g°C not 1.0 cal since the conversion is using grams and energy (J)

Answer 2

If you change that and repost, someone will check it.

Answer 3

You have to multiply it by the specific temperature of water